Equilateral triangle $ABC$ has been creased and folded so that vertex $A$ now rests at $A'$ on $\overline{BC}$ as shown.  If $BA' = 1$ and $A'C = 2,$ then find the length of crease $\overline{PQ}.$

[asy]
unitsize(1 cm);

pair A, Ap, B, C, P, Q;

A = 3*dir(60);
B = (0,0);
C = (3,0);
Ap = (1,0);
P = 8/5*dir(60);
Q = C + 5/4*dir(120);

draw(B--C--Q--P--cycle);
draw(P--Ap--Q);
draw(P--A--Q,dashed);

label("$A$", A, N);
label("$A'$", Ap, S);
label("$B$", B, SW);
label("$C$", C, SE);
label("$P$", P, NW);
label("$Q$", Q, NE);
[/asy]
Answer: The side length of equilateral triangle $ABC$ is 3.

Let $x = BP.$  Then $AP = A'P = 3 - x,$ so by the Law of Cosines on triangle $PBA',$
\[(3 - x)^2 = x^2 + 3^2 - 2 \cdot x \cdot 3 \cdot \cos 60^\circ = x^2 - 3x + 9.\]Solving, we find $x = \frac{8}{5}.$

Let $y = CQ.$  Then $AQ = A'Q = 3 - y,$ so by the Law of Cosines on triangle $QCA',$
\[(3 - y)^2 = y^2 + 2^2 - 2 \cdot y \cdot 2 \cdot \cos 60^\circ = y^2 - 2y + 4.\]Solving, we find $y = \frac{5}{4}.$

Then $AP = \frac{7}{5}$ and $AQ = \frac{7}{4},$ so by the Law of Cosines on triangle $APQ,$
\[PQ^2 = \sqrt{\left( \frac{7}{5} \right)^2  - \frac{7}{5} \cdot \frac{7}{4} + \left( \frac{7}{4} \right)^2} = \boxed{\frac{7 \sqrt{21}}{20}}.\]